In Peskin and Schroeder's book p 83, the authors said

At any fixed time $t_0$, we can of course expand $\phi$ in terms of ladder operators: $$\phi(t_0, \textbf{x})=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}} }}(a_{\mathbf{p}} e^{i\textbf{p.x}}+a_{\mathbf{p}} ^\dagger e^{-i\textbf{p.x}}) (1) $$ ... $e^{iH_0 (t-t_0)} \phi (t_0, \mathbf{x}) e^{-iH_0 (t-t_0)} \equiv \phi_I(t,\mathbf{x}).$...

Since we can diagonize $H_0$, it is easy to construct $\phi_I$ explicitly:

$\phi_I(t, \mathbf{x} ) = \int \frac{d^3p}{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}} } \left( a_{\mathbf{p}} e^{-ipx} + a^{\dagger}_{\mathbf{p}} e^{ipx} \right) |_{x^0 = t - t_0} $ (4.15)

To derive (4.15), I thought about $$ e^{i H_0 (t-t_0)} a_{\mathbf{p}} e^{-i H_0 (t-t_0)} $$ use BCH expansion in terms of $e^{-B} A e^B$ to work out (4.15), similar to the free field theories.

As answered in Interacting Fields in QFT $a_{\mathbf{p}}$ is time-indenpendent for free theory (I thought about $(\partial^2 + m^2) \phi(x) =0 $ in Eq. (2) in the aforementioned link for free theories) but time-dependent on interacting theories. Will this difference affect commutation relation $[a^{int}_{\mathbf{p}},H_0]$? More specifically, $$ [a^{int}_{\mathbf{p}},H_0 = \int \frac{d^3p }{(2\pi)^3} \omega_p a_{\mathbf{p}}^{free, \dagger} a_{\mathbf{p}}^{free} ] =? $$


1 Answer 1


By considering the Heisenberg-picture, we can deduce that given a free Hamiltonian $H_0$ and the annihilation operator $a_{\bf{k}}$, we have that $$[a_{{\bf{k}}},H_0]=\omega_{{\bf{k}}}a_{{\bf{k}}}\implies a_{\bf{k}}(t)=e^{iH_0t}a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{\bf{k}}t}a_{\bf{k}}.$$Here $a_{\bf{k}}(t)$ is the time dependent annihilation operator. Since $\omega_{\bf{k}}\in\mathbb{R}$, we have that $e^{-i\omega_{\bf{k}}t}\in\mathbb{C}$. We can use this to compute the commutation relation between $a_{\bf{k}}(t)$ and $H_0$. We thus have$$[a_{\bf{k}}(t),H_0]=[e^{-i\omega_{\bf{k}}t}a_{\bf{k}},H_0]=e^{-i\omega_{\bf{k}}t}[a_{\bf{k}},H_0]=\omega_{\bf{k}}e^{-i\omega_{\bf{k}}t}a_{\bf{k}}.$$But we have already established that $e^{-i\omega_{\bf{k}}t}a_{\bf{k}}=a_{\bf{k}}(t)$. Thus we have that$$[a_{\bf{k}}(t),H_0]=\omega_{\bf{k}}a_{\bf{k}}(t).$$ So we can see that the time-dependent annihilation (for example) commutates with $H_0$ in the same manner as the time-independent annihilation operator. Intuitively, this makes sense; if the Hamiltonian doesn't change with time, then at all times, the creation and annihilation operators will create/annihilate the same eigenstates $|k\rangle$ with corresponding energy eigenvalues $\omega_{\bf{k}}$.

It should be noted that this is not true in general. For a general Hamiltonian $H$, the commutation relations might be far more complicated. Furthermore, equal-time commutation relations can be shown to preserve the unitary structure of the operators and unequal-time commutation relations may vary.

  • $\begingroup$ Thanks. My concern is, the difference betwen $a^{int}_p$ and $a^{free}_p$, based on Eq. (2) in the answer in physics.stackexchange.com/questions/310353/…, is the non-zero part of $(\partial^2 + m^2)\phi(x) = H_{int} \phi(x)$. The term triggerd by $H_{int}$ can possibly be complicated. I am not sure if we can have $a_k(t) = e^{- i \omega_k t} a_k$ for $a_k^{int}$ $\endgroup$
    – AlphaF20
    Dec 28, 2021 at 20:59
  • $\begingroup$ $a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{\bf{k}}t}a_{\bf{k}}$ is derived from free field, I think. $\endgroup$
    – AlphaF20
    Dec 28, 2021 at 21:16
  • $\begingroup$ You can show that $e^{iH_0t}a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{{\bf{k}}}t}a_{{\bf{k}}}$. In the answer you referenced, it is shown that the time derivative of $a_{{\bf{k}}}$ is non-zero only when the Hamiltonian has an interacting component. The $a_{\bf{k}}$ (free) is constant in time when your theory is described only by a free field $H_0$. Whilst $a_{\bf{k}}(t)$ will in general depend on the interacting Hamiltonian and will not necessarily be the same for all time. $\endgroup$ Dec 29, 2021 at 3:00
  • $\begingroup$ That's my concern. I can derive $e^{iH_0t}a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{\bf{k}}t}a_{\bf{k}}$, in a way similar to P&S (2.46). By using $e^-B A e^B = A + [A, B] + \cdots$, noticing $[A,B] = [a_{\mathbf{k}}, -iHt] $, and P&S (2.32) $[H, a_{\mathbf{k}} ] = -\omega_{\mathbf{k}} a_{\mathbf{k}}$. But, that's all for free fields. My question is about if the above derivation is also valid in interacting theories, for specifically, $[a^{free}_{\mathbf{k}}, a^{int, \dagger}_{\mathbf{k}'}] = ?$. I'm not sure if I overlooked your point. $\endgroup$
    – AlphaF20
    Dec 29, 2021 at 4:53
  • $\begingroup$ Right, I see the confusion. Formally, the $e^{iH_0t}a_{\bf{k}}e^{-iH_0t}$ is called the interaction picture operator because when the coupling constant in $H_{\text{int}}$ is small (and usually it is assumed to be), most of the time dependence in $a_{\bf{k}}(t)$ will come from $H_0$, hence my previous arguments. If you want to describe the entire field, you then impose unitary operators that have $H$ and $H_0$ both involved. At that point I believe that closed form commutation relations seize to exist. $\endgroup$ Dec 29, 2021 at 6:50

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